HDU5008--Boring String Problem(SA+二分)

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最新推荐文章于 2024-08-13 10:26:25 发布

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分类专栏：字符串_后缀数组

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本文介绍了一种用于处理字符串查询的高效算法，通过构建后缀数组和高度数组来快速定位特定查询条件下的子串。该算法利用了二分查找和区间最小查询等技巧，实现了对大规模字符串的有效处理。

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Problem Description

In this problem, you are given a string s and q queries.

For each query, you should answer that when all distinct substrings of string s were sorted lexicographically, which one is the k-th smallest.

A substring s_i...j of the string s = a₁a₂ ...a_n(1 ≤ i ≤ j ≤ n) is the string a_ia_i+1 ...a_j. Two substrings s_x...y and s_z...w are cosidered to be distinct if s_x...y ≠ S_z...w

Input

The input consists of multiple test cases.Please process till EOF.

Each test case begins with a line containing a string s(|s| ≤ 10⁵) with only lowercase letters.

Next line contains a postive integer q(1 ≤ q ≤ 10⁵), the number of questions.

q queries are given in the next q lines. Every line contains an integer v. You should calculate the k by k = (l⊕r⊕v)+1(l, r is the output of previous question, at the beginning of each case l = r = 0, 0 < k < 2⁶³, “⊕” denotes exclusive or)

Output

For each test case, output consists of q lines, the i-th line contains two integers l, r which is the answer to the i-th query. (The answer l,r satisfies that s_l...r is the k-th smallest and if there are several l,r available, ouput l,r which with the smallest l. If there is no l,r satisfied, output “0 0”. Note that s_1...n is the whole string)

Sample Input

aaa
4
0
2
3
5

Sample Output

1 1
1 3
1 2
0 0
思路：每个sa[i]会贡献的新子串数为len-sa[i]-height[i];
     用前缀和搞出来，那么很容易求出一个l,r满足为所求子串（但是此时不是序号最小）
     这里有个地方要想清楚，序号的更新，这个串要么为sa[i]的一部分，要么为sa[>i]的一部分。不可能在sa[<i]。
     不然就不是sa[i]贡献的新子串了。。所以只需二分右端点。。最后就是RMQ求最小序号了。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;
#define maxn 240080
#define inf 0x3f3f3f3f
#define LL long long int
int str[maxn],vis[maxn];
char s[maxn];
int sa[maxn],t[maxn],t2[maxn],c[maxn],key[maxn];
int height[maxn],Rank[maxn];
int len1,len2;
int Min[maxn][20],lca[maxn][20];
LL dp[maxn];
inline int max(int a,int b)
{
    return a>b?a:b;
}
/*
用SA模板注意在最后添加一个比所有字符都小的字符。
key[n] = 0;
build_sa(key,n+1,m);
getHeight(key,n+1);
显然sa[0] 就是最后那个位置。。。
height[i] 表示 sa[i] 和 sa[i-1] 的最长公共前缀。。
*/

void build_sa(int * s,int n,int m)
{
    int i,*x = t,*y = t2;
    for(i = 0;i < m;i++)    c[i] = 0;
    for(i = 0;i < n;i++)    c[ x[i] = s[i] ]++;
    for(i = 1;i < m;i++)    c[i] += c[i-1];
    for(i = n-1;i >= 0;i--)    sa[--c[x[i]]] = i;
    for(int k = 1;k <= n;k <<= 1)
    {
        int p = 0;
        for(i = n - k;i < n;i++)    y[p++] = i;
        for(i = 0;i < n;i++)    if(sa[i] >= k)    y[p++] = sa[i] - k;
        for(i = 0;i < m;i++)    c[i] = 0;
        for(i = 0;i < n;i++)    c[ x[y[i]] ]++;
        for(i = 0;i < m;i++)    c[i] += c[i-1];
        for(i = n-1;i >= 0;i--)    sa[--c[x[y[i]]]] = y[i];
        //根据sa和y数组计算新的数y组
        swap(x,y);
        p = 1;    x[sa[0]] = 0;
        for(i = 1;i < n;i++)
            x[sa[i]] = y[sa[i-1]] == y[sa[i]] && y[sa[i-1] + k] == y[sa[i] + k] ? p-1:p++;
        if(p >= n)    break;
        m = p;
    }
}

void getHeight(int * s,int n)
{
    int i,j,k = 0;
    for(i = 0;i < n;i++)    Rank[sa[i]] = i;
    for(i = 0;i < n;i++)    
    {
        if(k) k--;
        int j = sa[Rank[i]-1];
        while(s[i+k] == s[j+k])        k++;
        height[Rank[i]] = k;
    }
}

void RMQ_INIT(int n)//求lca
{
    for(int i = 1;i < n;i++)    lca[i][0] = height[i];
	for(int j = 1;(1<<j)<=n;j++)
	{
		for(int i = 0;i+(1<<j)-1<n;i++)
		{
			lca[i][j] = min(lca[i][j-1],lca[i+(1<<(j-1))][j-1]);
		}
	}
}

int RMQ_Query(int l,int r)
{
    int k = 0;
    while((1<<(k+1) <= r-l+1)) k++;
    return min(lca[l][k],lca[r-(1<<k)+1][k]);
}
void RMQ_INIT1(int n)//求区间最小,这是求出sa后求得
{
    for(int i = 1;i < n;i++)    key[i] = sa[i]+1;
    for(int i = 1;i < n;i++)    Min[i][0] = key[i];
    for(int j = 1;(1<<j)<=n;j++)
	{
		for(int i = 0;i+(1<<j)-1<n;i++)
		{
			Min[i][j] = min(Min[i][j-1],Min[i+(1<<(j-1))][j-1]);
		}
	}
}
int RMQ_Query1(int l,int r)//
{
    int k = 0;
    while((1<<(k+1)) <= r-l+1) k++;
    return min(Min[l][k],Min[r-(1<<k)+1][k]);
}
int main()
{
   // freopen("in.txt","r",stdin);
    while(scanf("%s",s)!=EOF)
    {
        int len = strlen(s);
        for(int i = 0;i < len;i++)
            str[i] = s[i]-'a'+2;
        str[len] = 0;
        build_sa(str,len+1,30);
        getHeight(str,len+1);
        dp[0] = 0;
        dp[1] = len-sa[1];
        for(int i = 2;i <= len;i++)
        {
            LL add = len-sa[i]-height[i];
            dp[i] = dp[i-1] + add;
        }
        RMQ_INIT(len+1);
        RMQ_INIT1(len+1);
        LL l = 0,r = 0,v;
        int q;    scanf("%d",&q);
        while(q--)
        {
            scanf("%I64d",&v);
            LL k = (l^r^v)+1;
            if(k > dp[len])
            {
                l = r = 0;
                cout << 0 << " " << 0 << endl;
                continue;
            }
            int pos = lower_bound(dp+1,dp+1+len,k)-dp;
            k -= dp[pos-1];
            int L = sa[pos];
            int R = L+k+height[pos]-1;////非常没问题
            //这样从L 到 R就是所要求的字符串，但是还不是满足最小的序号的
            int Len = R-L+1;
            int ll = pos+1,rr = len;
            int ans = pos;
            while(ll <= rr)
            {
                int mid = (ll+rr)>>1;
                if(RMQ_Query(pos+1,mid) < Len) rr = mid-1;
                else 
                {
                    ans = mid;
                    ll = mid+1;
                }
            }
            int fuck = RMQ_Query1(pos,ans);
            l = fuck,r = fuck+Len-1;
            printf("%I64d %I64d\n",l,r);
        }
    }
    return 0;
}